I shall write the proof of this following theorem, which i have found on the internet. i shall mark the part which i do not understand.

The proof is done by contradiction:

Let $B$ be not connected, thus there exists open, disjunct none empty sets $U, V$ such that $B \subset U \cup V $ with $B \cap U$ and $B \cap V$ non-empty.
Because $ A \subset B $ then obviously $A \subset U \cup V $.
Furthermore are the intersections $ A \cap V $ and $ A \cap U $ none empty. For if for example $ A \cap U = \emptyset $ then $A \subset X-U$ then because $ X - U $ is closed (Because $U$ is open) we have

  1. $ B \subset \bar{A} \subset X-U$

This is however is a contradiction to the fact that $B \cap U \neq \emptyset $

Analog we find that $ A \cap V$ is not empty

But then $ V , U $ would disassemble A then A would not be connected which is a contradiction.

What i do not understand is marked as equation (1). Why does $\bar{A} \subset X-U$ ? The rest of the proof is clear to me. Thanks for your help.


The closure $\overline{A}$ of $A$ in $X$ is the smallest closed subset of $X$ containing $A$. More precisely, it is the intersection of all closed subsets of $X$ containing $A$ and thus contained in every closed subset containing $A$. In your case the set $X - U$ is closed so that $A \subset X - U$ implies $\overline{A} \subset X - U$.

  • $\begingroup$ Interesting, does Generally speaking $ A \subset B $ where as B closed thus we write $ A \subset B = \bar {B} $ imply that $ \bar{A} \subset B = \bar {B}$ ? $\endgroup$
    – Mad Spaces
    May 29 at 10:40
  • 2
    $\begingroup$ @MadSpaces Yes. $\endgroup$
    – mrtaurho
    May 29 at 10:40
  • $\begingroup$ Thank your for explaining. $\endgroup$
    – Mad Spaces
    May 29 at 10:41
  • 3
    $\begingroup$ @MadSpaces : Maybe even a bit more general, $A \subset B$ implies $\overline{A} \subset \overline{B}$ for any $A$ and $B$. $\endgroup$ May 29 at 10:44
  • $\begingroup$ You are right @MatthiasKlupsch. Thanks $\endgroup$
    – Mad Spaces
    May 29 at 12:36

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